Problem 23 - Thomas Calculus

Question

Evaluate \(\int_{C} \frac{x^{2}}{y^{4 / 3}} d s,\) where \(C\) is the curve \(x=t^{2}, y=t^{3},\) for \(1 \leq t \leq 2\)

Step-by-Step Solution

Answer

The integral transforms to \( \int_{1}^{2} t^2 \sqrt{4 + 9t^2} \, dt \).

Step 1: Understanding the Problem

We need to evaluate the line integral \( \int_{C} \frac{x^{2}}{y^{4/3}} \, ds \), where the curve \( C \) is parameterized by \( x(t) = t^{2} \), \( y(t) = t^{3} \) for \( 1 \leq t \leq 2 \). To solve this, we can substitute the parametrization into the integral.

Step 2: Substitute Parametric Equations

Substitute \( x = t^2 \) and \( y = t^3 \) into the integral to get \( \frac{(t^2)^2}{(t^3)^{4/3}} \). Simplifying gives us \( \frac{t^4}{t^4} = 1 \). The problem now reduces to finding the arc length \( ds \).

Step 3: Find the Differential Arc Length Element

The arc length differential \( ds \) is given by \( ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \). Calculate \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = 3t^2 \) to find \( ds = \sqrt{ (2t)^2 + (3t^2)^2 } \, dt = \sqrt{4t^2 + 9t^4} \, dt \).

Step 4: Evaluate the Integral

Substitute \( ds = \sqrt{4t^2 + 9t^4} \, dt \) into the integral. The integral becomes \( \int_{1}^{2} 1 \times \sqrt{4t^2 + 9t^4} \, dt \). Simplifying \( \sqrt{4t^2 + 9t^4} = t^2 \sqrt{4 + 9t^2} \) reduces the integral to \( \int_{1}^{2} t^2 \sqrt{4 + 9t^2} \, dt \), which requires a substitution or other method to solve.

Key Concepts

Parametric Equations

Parametric equations are a way to represent a curve by using one or more parameters, typically denoted as "t." In the given problem, the curve \( C \) is defined by the parametric equations \( x(t) = t^2 \) and \( y(t) = t^3 \). This means that, as the parameter \( t \) changes from 1 to 2, both \( x \) and \( y \) change according to these functions.
This parameterization serves several purposes in calculus and geometry:

• Flexibility: Parametric equations can easily describe curves that cannot be represented as functions \( y=f(x) \) or \( x=g(y) \).
• Dynamic Representation: Allows tracing out the path and orientation of a curve as the parameter changes.
• Simplified Calculations: Makes it easier to calculate derivatives and integrals along a curve.

In this exercise, the parametric form simplifies the process of expressing and computing the integral by substituting \( t \) directly into the equations for \( x \) and \( y \). This substitution reflects the inherent relationship between \( x \) and \( y \) as they change along the path \( C \).

Differential Arc Length

When dealing with curves, especially in the context of line integrals, understanding the concept of differential arc length is crucial. The differential arc length, \( ds \), represents an infinitesimally small segment of the curve, analogous to the way \( dx \) or \( dy \) represents tiny segments of straight lines in standard cartesian coordinates.
The formula to compute this differential \( ds \) for a parameterized curve \( x(t) \) and \( y(t) \) is:

• \( ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \)

In the given problem, substituting the derivatives \( \frac{dx}{dt} = 2t \) and \( \frac{dy}{dt} = 3t^2 \) into the formula yields:

• \( ds = \sqrt{(2t)^2 + (3t^2)^2} \, dt = \sqrt{4t^2 + 9t^4} \, dt \)

Calculating \( ds \) allows us to weigh the contributions of each infinitesimal element of the curve towards the overall integral. In this specific exercise, finding the correct expression for \( ds \) makes the evaluation of the integral manageable.

Substitution Method

The substitution method is a powerful technique in calculus used to simplify integrals and make them solvable either analytically or numerically. In the context of line integrals, substitution often involves substituting parametric equations into the integral.
Once a parametric equation is substituted, as in our problem where \( x = t^2 \) and \( y = t^3 \), the integral becomes:

• \( \int_{C} \frac{(t^2)^2}{(t^3)^{4/3}} \, ds = \int_{1}^{2} 1 \times \sqrt{4t^2 + 9t^4} \, dt \)

This substitution simplifies the original problem by reducing it to a more straightforward form. Instead of handling the complexities of the original expressions, the substitutions condense the integral into functions of \( t \) that are often easier to solve.
While the substitution method is essential for the integration process, it can be complemented by other methods like numerical approximation when exact solutions are complicated or impossible to derive. Understanding this method provides a critical framework for solving complex integrals encountered in calculus.